If $|x|+|y|>|x+z|$, then is $y > z, y<z$, or $y=z$, or not determinable?

Question

If $|x|+|y|>|x+z|$, then is $y > z, y<z$, or $y=z$, or not determinable?

I thought about cases, $x>0,x<0,y>0,y<0,z>0,z<0.$ Plugging numbers is unconvincing to me.

Answer 1

In a simpler mathematical way without using numbers, you can see that this result is indeterminable.

Note that $\forall a,b\in\mathbb R$, we have

$$|a|+|b|≥|a+b|$$

Hence putting $y=z\neq 0$ and $y=-z\neq 0$ with some $x$, such that the inequality $|x|+|z|>|x+z|$ is always correct. This implies that even at least one of the results $y>z$ or $y<z$ or $y=z$ cannot always be correct.

Answer 2

Here are simple approaches that don't use numbers:

• 1. Putting $x=-z$ , we get

$$|y|+|z|>0$$

Then, you can choose infinitely many pair $y,z\neq 0$, such that all possible cases: $y>z$ or $y<z$ or $y=z$ hold.

• 2. Fix $x=0$, you get $|y|>|z|$. This implies, you can also use the same argument. But, note that in this case $y=z$ doesn't hold.